University of Florida/Egm4313/s12.team11.R5

Report 5


Intermediate Engineering Analysis
Section 7566
Team 11
Due date: March 30, 2012.

R5.1Edit

Problem StatementEdit

Given: Find   for the following series:

1.  
2.  

Find   for the Taylor series of

3.   at  

4.   at  

5.   at  

SolutionEdit

The radius of convergence   is defined as

 

1.  
 
 
 
 

                                                           

2.  
 
 
 
 

However, in this problem, the series   term is   not  , as is the general form.
Therefore, this implies:

 
 

                                                         


3. The Taylor series for   is expressed as  

 

 

 

 

Therefore:  

                                                            

4. The Taylor series for   at   is expressed as  

 

 

 

 

 

                                                              



5. The Taylor series for   at   is expressed as  

 

 

 

 

 

For convergence:  

 

 

Therefore,

                                                              



R5.2Edit

Solved by: Andrea Vargas

Problem StatementEdit

Part 1:Determine whether the following are linearly independent using the Wronskian

Part 2: Determine whether the following are linearly independent using the Gramian

SolutionEdit

Part 1Edit

Using the Wronskian we check for linear independence.


We know from (1) and (2) in 7-35 that if
 
Then the functions are linearly independent.

Part 1.1Edit

 
 

Taking the derivatives of each function:


 
 


 

                                            f(x) and g(x) are linearly independent
Part 1.2Edit

 
 

Taking the derivatives of each function:


 
 


 

                                             f(x) and g(x) are linearly independent

Part 2Edit

Using the Gramian we check for linear independence.
We know from the notes in (1) 7-34 that:
 

and that the Gramian is defined as:  

Then f,g are linearly independent if  

Part 2.1Edit

 
 

Taking scalar products:
 
 
 

 

                                            f(x) and g(x) are linearly independent
Part 2.2Edit

 
 

Taking scalar products:
 
We can use the trig identity for power reduction  
Then we have,
 
 

 
 
 

 
 
 


 

                                            f(x) and g(x) are linearly independent

ConclusionEdit

By both methods (the Wronskian and the Gramian) we obtain the same results.

R5.3Edit

Problem StatementEdit

Verify using the Gramian that the following two vectors are linearly independent.

 
 

SolutionEdit

 

We know from (3) 8-9 that:
 

We obtain,
 
   

Then,
 

                                             b_1 and b_2 are linearly independent

R5.4Edit

Problem StatementEdit

Show that   is indeed the overall particular solution of the L2-ODE-VC   with the excitation  .

Discuss the choice of   in the above table e.g., for   why would you need to have both   and   in   ?

SolutionEdit

Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:

  is a specific excitation with known form of   and   is a specific excitation with known form of  

 

becomes

 

proving that

  is indeed the overall particular solution of the L2-ODE-VC   with the excitation  


According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:

 

where the coefficients   are the Fourier coefficients calculated using Euler formulas.

So even though the system is being excited by functions like   the particular solution would still include both   and   in   because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both   and   times the Fourier coefficients

R5.5Edit

Part 1Edit

Problem StatementEdit

Show that   and   are linearly independant using the Wronskian and the Gramain (integrate over 1 period)

SolutionEdit

 
One period of  
Wronskian of f and g
 

Plugging in values for  
   
 
 

 They are linearly Independant using the Wronskian.

 
 
 
 
 
 
 

 They are linearly Independent using the Gramain.

Problem StatementEdit

Find 2 equations for the 2 unknowns M,N and solve for M,N.

SolutionEdit

 
 
 
Plugging these values into the equation given ( ) yields;
 
Simplifying and the equating the coefficients relating sin and cos results in;
 
 
Solving for M and N results in;

   

Problem StatementEdit

Find the overall solution   that corresponds to the initial conditions  . Plot over three periods.

SolutionEdit

From before, one period   so therefore, three periods is  
Using the roots given in the notes  , the homogenous solution becomes;
 
Using initial condtion  ;
 
 
with  
 
Solving for the constants;
 
 
Using the   found in the last part;
 

  

 

 

R5.6Edit

solved by Luca Imponenti

Problem StatementEdit

Complete the solution to the following problem

 

where

 

and

 

Find the overall solution   corresponds to the initial condition:

 

Plot the solution over 3 periods.

Particular SolutionEdit

Taking the derivatives of the particular solution  

 

 

 

Plugging these into the ODE yields

     

Equating like terms allows us to solve for M and N

 

 

 

 

 

So the particular solution is

 

Overall SolutionEdit

The overall solution in the sum of the homogeneous and particular solutions

 

 

To find A and B we apply the initial conditions

 

 

Taking the derivative

 

 

 

 

Giving us the overall solution

    

PlotEdit

The period for   is  

Plotting the solution   over 3 periods yields

 

R5.7Edit

Solved by Daniel Suh

Problem StatementEdit

 

 

 

1. Find the components   using the Gram matrix.
2. Verify the result by using   and  , and rely on the non-zero determinant matrix of   and   relative to the bases of   and  .

Part 1 SolutionEdit

Gram MatrixEdit

 

 
Thus,

 

 

 

 

 


Defining cEdit

Define:  
 

If  , then   exists

 


Finding cEdit

  thus,  exists

 

 

                                                      

Part 2 SolutionEdit

 

 

 

 

 

                                                       solution is correct

R5.8Edit

Problem StatementEdit

Find the integral

  for   and  

Using integration by parts, and then with the help of of

General Binomial Theorem

 

SolutionEdit

For  :

 

For substitution by parts,  

 

 

 

Therefore:

                                      

Using the General Binomial Theorem:

 

Therefore:  

Which we have previously found that answer as:

                                      




For  :

 

Initially we use the following substitutions:  

 

First let us consider the first term:  

Next, we use the integration by parts:  
 

 

 

Next let us consider the second term:  

Again, we will use integration by parts:  
 

 

 

Therefore:

 

 

Re-substituting for t:

 

 

 

Therefore:

                                      



Using the General Binomial Theorem for the integral with t substitution  :

 

Therefore:  

Which we have previously found that answer as:

                                      

R5.9Edit

Solved by: Gonzalo Perez

Problem StatementEdit

Consider the L2-ODE-CC (5) p.7b-7 with   as excitation:

  (5) p.7b-7

  (1) p.7c-28

and the initial conditions

 .

Part 1Edit

Part AEdit

Project the excitation   on the polynomial basis

  (1)

i.e., find   such that

  (2)

for x in  , and for n = 3, 6, 9.

Plot   and   to show uniform approximation and convergence.

Note that:

  (3)

SolutionEdit

To solve this problem, it is important to know that the scalar product is defined as the following:

 .

Therefore, it follows that:

 , where   and  .

We know that if   are linearly independent, then by theorem on p.7c-37, the matrix is solvable.

According to this and (3)p.8-14:

If   exists  . (3)p.8-14

Now let's define the Gram matrix   as a function of  :

  (1)p.8-13

Defining the "d" matrix as was done in (3)p.8-13, we get:

 . (3)p.8-13

And according to (1)p.8-15:   (1)p.8-15

Now, we can find the values to compare   to  .

Using Matlab, this is the code that was used to produce the results:

File:Matlab59.JPG

The Matlab code above produced the following graph:

File:Matlab59graph.jpg

Where   is represented by the dashed line and the approximation, , is represented by the red line. This code can work for all n values.

Part BEdit

In a seperate series of plots, compare the approximation of the function   by Taylor series expansion about  .

Where:  

SolutionEdit

For n=1:

 

For n=2:

 

For n=3:

 

For n=4:

 

For n=5:

 

For n=6:

 

For n=7:

 

For n=8:

 

For n=9:

 

For n=10:

 

For n=11:

 

For n=12:

   

For n=13:

   

For n=14:

   

For n=15:

   

For n=16:

   

Using Matlab to plot the graph:

File:4.3Agraph.jpg

Part 2Edit

Find   such that:

  (1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot   for n = 3, 6, 9, for x in  .

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

SolutionEdit

First, we find the homogeneous solution to the ODE:
The characteristic equation is:
 
 
Then,  
Therefore the homogeneous solution is:
 

Now to find the particulate solution

For n=3:

 

 

We can then use a matrix to organize the known coefficients:

 

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 

Superposing the homogeneous and particulate solution we get
 

Differentiating:
  Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally we have:
 

For n=6:

 

 

We can then use a matrix to organize the known coefficients:

 

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 

Superposing the homogeneous and particulate solution we get
 

Differentiating:
  Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally
 

For n=9:

 

 

We can then use a matrix to organize the known coefficients:
   

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 
 

Superposing the homogeneous and particulate solution we get
 
 

Differentiating:
   
Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally
 
 

Here is the graph for this problem using Matlab:

File:4.3Bgraph.jpg